Answer to Question #149135 in Trigonometry for Al-Amin Akash

Question #149135

"1+4x^2 = 4x.secA"

Sec6A - tan6A =?

1
Expert's answer
2020-12-08T16:58:18-0500

Given,


"1+4x^2=4xsecA" , so write "secA" in terms of "x" as "secA=\\frac{1+4x^2}{4x}"




Use trigonometry to find "p" in terms of "x" as,


"p=\\sqrt{(1+4x^2)^2-(4x)^2}"


"=\\sqrt{1+16x^4+8x^2-16x^2}"


"=\\sqrt{1+16x^4-8x^2}"


"=\\sqrt{1^2+(4x^2)^2-2(1)(4x^2)}"


"=\\sqrt{(1-4x^2)^2}"


"=1-4x^2"


So, "tanA=\\frac{1-4x^2}{4x}"


Now, simplify the expression "sec^6A-tan^6A" as,


"sec^6A-tan^6A=(sec^2A)^3-(tan^2A)^3"


Use formula "a^3-b^3=(a-b)(a^2+ab+b^2)" to obtain,


"sec^6A-tan^6A=(sec^2A-tan^2A)(sec^4A+sec^2Atan^2A+tan^4A)"


"=(1)(sec^4A+sec^2Atan^2A+tan^4A)"


"=sec^4A+sec^2Atan^2A+tan^4A"


"=sec^4A+sec^2Atan^2A+(sec^2A-1)^2"


"=sec^4A+sec^2Atan^2A+sec^4A-2sec^2A+1"


"=2sec^4A+sec^2Atan^2A-2sec^2A+1"


"=2sec^2A(sec^2A-1)+sec^2Atan^2A+1"


"=2sec^2Atan^2A+sec^2Atan^2A+1"


"=3sec^2Atan^2A+1"


Substitute "secA=\\frac{1+4x^2}{4x}" and "tanA=\\frac{1-4x^2}{4x}" into simplified expression to obtain,


"=3(\\frac{1+4x^2}{4x})^2(\\frac{1-4x^2}{4x})^2+1"


"=3[(\\frac{1+4x^2}{4x})(\\frac{1-4x^2}{4x})]^2+1"


"=3[\\frac{(1+4x^2)(1-4x^2)}{16x^2}]^2+1"


"=3[\\frac{1-16x^4}{16x^2}]^2+1"


"=3(\\frac{1+256x^8-32x^4}{256x^4})+1"


"=\\frac{3+768x^8-96x^4+256x^4}{256x^4}"


"=\\frac{768x^8+160x^4+3}{256x^4}"


Therefore, the simplified expression in terms of "x" is "sec^6A-tan^6A==\\frac{768x^8+160x^4+3}{256x^4}"



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